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Gekata [30.6K]
3 years ago
9

For Madeline's lemonade recipe, 3 lemons are required to make 6 cups

Mathematics
1 answer:
expeople1 [14]3 years ago
6 0

Answer:

1/2 lemon per 1 cup

Step-by-step explanation:

6/6=1

3/6=1/2= .5 lemon

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I have to find the area of this rectangular prism pls help!!
Nina [5.8K]

Answer:

the surface area is 208

Step-by-step explanation:

formula for surface area is A = 2 (w*l + h*l + h*w )

A = 2 ( 4*8 + 6*8 + 6*4 )

A = 2 ( 32 + 48 + 24 )

A = 2 ( 104 )

A = 208

8 0
3 years ago
1.Give at least two characteristics for the rhombus.
jok3333 [9.3K]
1. opposite sides are equivalent , opposite angles are equivalent
2. 4 sides, 4 angles, both answers for number one also work
3. A square
5 0
3 years ago
If 5 burgers and 4 orders of fries cost $30.76 and 8 burgers and 6 orders of fries cost $48.28, what is the cost of a burger and
padilas [110]

Step-by-step explanation:

5B+4F= $30.76

×3

15B+12F= $92.28

----------------------------

8B+6F= $48.28

×2

16B+12F= $96.56

----------------------------

Cost of one burger:

$97.56-$92.28= $4.28 (final ans)

Cost of one fries:

$30.76-5($4.28)= $9.36 (for 5 fries)

$9.36÷5= $1.87 (final ans)

3 0
3 years ago
Choose the appropriate transformation for the figure<br> Picture above ***
kirill115 [55]
Answer :

Rotation :)
6 0
3 years ago
A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Kar
ch4aika [34]

Answer:

We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.8 minutes with a standard deviation of 2.5 minutes.

Let \mu = <u><em>mean waiting time for bus number 14.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 10 minutes      {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, H_A : \mu < 10 minutes    {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                       T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean waiting time = 7.8 minutes

             s = sample standard deviation = 2.5 minutes

             n = sample of different occasions = 18

So, <u><em>test statistics</em></u> =  \frac{7.8-10}{\frac{2.5}{\sqrt{18} } }  ~ t_1_7

                              =  -3.734

The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean waiting time is less than 10 minutes.

5 0
3 years ago
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